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    center of gravity

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    Rotational Equilibrium - a body is in rotational equilibrium when no net torque acts on it. The sum of the torques is equal to zero. Conditions for Equilibrium The first condition of equilibrium: “For a body to be in equilibrium‚ the vector sum of all the forces acting on that body must be zero.” (ΣF = 0) The second condition of equilibrium: “For a body to be in rotational equilibrium‚ the sum of all the torques acting on that body must be zero.” (ΣΓ = 0)

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    srm motor

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    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Motor Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.1 Torque-Speed Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2 Electromagnetic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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    Simulink Implementation of Induction Machine Model – A Modular Approach Burak Ozpineci1 Leon M. Tolbert1‚2 burak@ieee.org tolbert@utk.edu 1 2 Oak Ridge National Laboratory P.O. Box 2009 Oak Ridge‚ TN 37831-6472 Department of Electrical and Computer Engineering The University of Tennessee Knoxville‚ TN 37996-2100 parameters are accessible for control and verification purposes. Simulink induction machine model discussed in this paper has been featured in a recent graduate

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    Quiz on Power and Torque

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    1. What is a simpler way to expressing 0.000 018 A? 2. What is a simpler way to expressing 5 000 000 V? 3. An electric motor is developing 8.5kW at a speed of 900r/min. Calculate the torque available at the shaft. 4. A generating station has a daily output of 160MW h and uses 300 t (tones) of coal in the process. The coal releases 7 MJ/kg when burnt. Calculate the overall efficiency of the station. 5. A lift of 250 kg mass is raised with a velocity of 6 m/s. If the driving motor has an efficiency

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    Phy-1

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    the mass of the rotating body is distributed about its axis of rotation. This quantity is known as moment of inertia. We may now write‚ [pic] [pic] Thus‚ K. E = [pic] TORQUE: The

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    A SELF-BALANCING QUADCOPTER DESIGN WITH AUTONOMOUS CONTROL H. S. M. M. Caldera1‚ B. W. S. Anuradha1‚ D. M. G. K. P. Udgeethi1‚ A. A. T. Surendra1‚ B. R. Y. Dharmarathne1‚ R. D. Ranaweera2‚ D. Randeniya2 1Department of Mechatronics Engineering‚ South Asian Institute of Technology and Medicine‚ Sri Lanka‚ Email:shehanmalaka.c@gmail.com 2Department of Electrical and Electronic Engineering‚ University of Peradeniya‚ Sri Lanka Email:rdbranaweera@ee.pdn.ac.lk Abstract The unmanned aerial vehicles

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    Trifilar Suspension

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    body to angular acceleration [1]. An important factor as the resulting moment governs the analysis of rotational dynamics with an equation of the form M=I∝ which defines a relationship between several properties including angular acceleration and torque [2]. The polar moment of inertia is the measure of a body’s resistance to torsion and is used to calculate the angular displacement and periodic time of the body under simple harmonic motion [3]. The moment of inertia of any mechanical component that

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    Hydrostatic Force

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    any point on the curved surface is normal to the surface and therefore resolves through the pivot point because this is located at the origin of the radii. Hydrostatic forces on the upper and lower curved surfaces therefore have no net effect – no torque to affect the equilibrium of the assembly because all of these forces pass through the pivot. The forces on the sides of the quadrant are horizontal and cancel out (equal and opposite).

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    sin(α)α2 − mLr cos(α)¨ ˙ α ˙ = T − Bθ 4 2 ¨ ¨ 3 mL α − mLr cos(α)θ − mgL sin α = 0 2 Defining x= θ α ˙ θ α ˙ T ‚ y= θ α T ‚ u=V (1) where T is the torque on the load from the motor‚ α is the pendulum angle‚ θ is the horizontal arm angle and other system parameters are given in Table I. In addition‚ the torque T is generated by DC motor such that [2] T = ηm ηg Kt Kg ˙ V − Kg Km θ Rm (2) and linearizing about the upright position i.e. x = 0‚ yields     0 0 0 1 0  0 0 

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    Physics

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    Introductory Physics I Elementary Mechanics by Robert G. Brown Duke University Physics Department Durham‚ NC 27708-0305 rgb@phy.duke.edu Copyright Notice Copyright Robert G. Brown 1993‚ 2007‚ 2013 Notice This physics textbook is designed to support my personal teaching activities at Duke University‚ in particular teaching its Physics 141/142‚ 151/152‚ or 161/162 series (Introductory Physics for life science majors‚ engineers‚ or potential physics majors‚ respectively). It is freely

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